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Calculating stress and strains for beam of applied force

  1. Mar 28, 2016 #1
    1. The problem statement, all variables and given/known data
    If I consider a force, which is applied to both ends of a rectangular beam to its cross section with dimensions w (width) and h (height) and I know the length of the beam is l and the force is parallel, how can I calculate the stress?

    Also, what if the force were parallel to the [111] direction? How could I calculate the stress and strain considering the force is not parallel?

    2. Relevant equations
    For the first part, I think I would just use:
    $$\sigma = F/A = \frac{F}{wh}$$

    I now that A is the area perpendicular to the applied force.
    But do I need to anything specific to account for the fact that it is applied at both ends?

    Also, when the force applied is in the [111] direction, I am not sure how to calculate the stress and strain.




    3. The attempt at a solution
    As per the relevant equations, I show my attempt. I just use
    $$\sigma = F/A = \frac{F}{wh}$$
    for the first part. Not sure if it should be 2F since the force is applied on both ends?

    I am not sure about how to calculate stress and strain when the force is parallel to the [111] direction. Any ideas? How do I calculate the area perpendicular to the applied force in [111] direction (just in variables).
     
  2. jcsd
  3. Mar 28, 2016 #2

    PhanthomJay

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    When you apply a force F perpendicular to the cross section (parallel to the axial length of the beam) , the reaction force at the other end must be F in the opposite direction, from equilibrium considerations. Thus, the net external force on the beam is zero. But, the internal force in the beam is non-zero, and that internal force perpendicular to the cross section is the force to use when determining the axial stress. So if you draw a free body diagram cutting through the beam, is that internal force F or 2F?
    For the second part, I don't know what you mean by the [111] direction, you mean perpendicular to the length? That's a stress of a different nature.
     
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